# Chapter 7 - Newton's Third Law - Exercises and Problems - Page 187: 25

The mug does not slide.

#### Work Step by Step

We can find the maximum possible magnitude of acceleration of the mug which can be provided by the force of static friction. $F_f = ma$ $mg~\mu_s = ma$ $a = g~\mu_s$ $a = (9.80~m/s^2)(0.50)$ $a = 4.90~m/s^2$ We can find the deceleration of the car during the braking period. $a = \frac{v^2-v_0^2}{2x}$ $a = \frac{0-(20~m/s)^2}{(2)(50~m)}$ $a = -4.0~m/s^2$ Since the mug's maximum possible magnitude of acceleration provided by the force of static friction is greater than the magnitude of deceleration of the car, the mug does not slide.

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